Is the 4D Volume-Time Element Invariant Under Lorentz Boost in the z Direction?

[Unown]

Homework Statement

Prove that the element dt\ dx\ dy\ dz is invariant under Lorentz boost with velocity \beta along z axis.

Homework Equations

Convention c=1

Lorentz boost in z direction:
L(z)=\left[ \begin{array}{cccc} \gamma & 0 & 0 & -\gamma\beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma\beta & 0 & 0 & \gamma \end{array} \right],\ \gamma=\frac{1}{\sqrt{1-\beta^2}}

Definition of invariance:
dt'dx'dy'dz'=dt\ dx\ dy\ dz

The Attempt at a Solution

Looks simple.
\left[ \begin{array}{c} dt' \\ dx' \\ dy' \\ dz' \end{array} \right]=\left[ \begin{array}{cccc} \gamma & 0 & 0 & -\gamma\beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma\beta & 0 & 0 & \gamma \end{array} \right]\left[ \begin{array}{c} dt \\ dx \\ dy \\ dz \end{array} \right]

dt'=\gamma dt-\gamma\beta dz,\ dx'=dx,\ dy'=dy,\ dz'=-\gamma\beta dt+\gamma dz

dt'dx'dy'dz'=\gamma^2(dt-\beta dz)\ dx\ dy\ (dz-\beta dt)

I got stuck, but then I noticed that \beta=dz/dt, so:

dt'dx'dy'dz'=\gamma^2 dt\left(1-\beta \frac{dz}{dt}\right)\ dx\ dy\ dz\left(1-\beta \frac{dt}{dz}\right)=\frac{1}{1-\beta^2} dt\left(1-\beta^2\right)dx\ dy\ dz\left(1-\frac{dz}{dt}\frac{dt}{dz}\right)=\left(1-1\right)\ dt\ dz\ dy\ dz=0

It shouldn't be zero! Please help!

 

[Dick]

You want a determinant of the matrix there don't you? Isn't that the Jacobian?

 

[Matterwave]

Uh...what happened to your dx on the right hand side?

 

[Unown]

Yes, if we use Jacobian this becomes trivial.
But the question doesn't include Jacobian, we have to do this the normal way.

I think my problem can be somewhere when I am writing \frac{dz}{dt} and \frac{dt}{dz}. Maybe there should be metric tensor somewhere that gives additional minus sign?

Please help!

 

[Matterwave]

I think that you wrote one of the dx's as dz...

Your answer came out quadratic in dz.

But perhaps that doesn't fix it, let me see...

 

[Unown]

Yes, sorry about that. I corrected this issue. But the equation still gives 0.

 

[Matterwave]

I'm actually not seeing how you're going from your third to last expression to your second to last expression...but I expanded out the thing myself and got 0 also, lol...sorry I guess I don't know where the error is.

 

[Unown]

In the second to last expression, you just take dt' dx' dy' and dz' from the third and place 'em there. I also took out \gammas, to simplify things.

Well thanks for the try. Just be sure you didn't get suggested by my derivation and did the same mistake.

Any other ideas?

 

[qbert]

given a mapping f: (x,y) -> (s,t)

ds dt is given by J(f) dx dy where J(f) is the Jacobian of the map.

it is NOT given by
((ds/dx) dx + (ds/dy)dy)*( (dt/dx)dx + (dt/dy)dy)
which is what you're trying to calculate above.

 

[Unown]

Okey. But why is that? Why is my calculation wrong?

How to show, that ds dt is given by J(f) dx dy?

 

[qbert]

Your calculation is wrong because that's not how volumes work! Why is the formula i wrote down right? Any book talking about vector calculus should explain it. But I'll give it a go.

Here's a sketch:
we need to be clear about what you mean by ds dt! from a geometric point of view
what we're really doing (in 2d) is calculating the AREA dA of the image of the map.

if the point (x,y) goes to f(x,y) = (s(x,y), t(x,y) ) = (s,t)

the point (x+dx, y) goes to f(x+dx, y) = (s(x+dx, y), t(x+dx, y) ) which is about
(s, t) + (ds/dx, dt/dx) dx

similarly (x, y+dy) goes to (s,t) + (ds/dy, dt/dy)dy

now dA is the area spanned by the two vectors
v1 = f(x+dx, y) - f(x,y) = (ds/dx, dt/dx) dx
v2 = f(x, y+dy) - f(x,y) = (ds/dy, dt/dy) dy
the area is just the (magnitude of the) cross product.

dA = |v1 X v2| = | (ds/dx, dt/dx) X (ds/dy, dt/dy) | dx dy
dA = |ds/dx dt/dy - ds/dy dt/dx| dx dy

so we write down
dA = dx dy = J(f) ds dt where J(f) is |ds/dx dt/dy - ds/dy dt/dx| the Jacobian of the transform.

----------------------------------------
now. Why doesn't your way work.
Let's look at a simple example:
s = 1/sqrt(2) ( x + y)
t = 1/sqrt(2) (-x + y)
(s and t here are just rotated by 90 degrees)

ds = 1/sqrt(2)( dx + dy)
dt = 1/sqrt(2)( -dx + dy)
ds dt =?= 1/2(dx + dy)(-dx + dy) = 1/2(-dx^2 +dx dy - dy dx + dy^2)
= 1/2(dy^2 - dx^2)
well now this is just wrong. there's nothing to be done here the differentials can't meaningfully be treated as numbers to work out volume changes.

what should we have gotten
ds dt = | 1/sqrt(2) 1/sqrt(2) - (-1/sqrt(2))(1/sqrt(2))| dx dy = dx dy

 

[Unown]

Thanks! I understand now.